Difference between revisions of "2019 AIME I Problems/Problem 2"
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− | The | + | ==Problem 2== |
+ | Jenn randomly chooses a number <math>J</math> from <math>1, 2, 3,\ldots, 19, 20</math>. Bela then randomly chooses a number <math>B</math> from <math>1, 2, 3,\ldots, 19, 20</math> distinct from <math>J</math>. The value of <math>B - J</math> is at least <math>2</math> with a probability that can be expressed in the form <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | The probability that <math>B-J<0</math> is <math>\frac{1}{2}</math> by symmetry. | ||
+ | The probability that <math>B-J= 1</math> is <math>\frac{19}{20 \times 19} = \frac{1}{20}</math> because there are 19 pairs: <math>(B,J) = (2,1),.., (20,19)</math>. | ||
+ | |||
+ | The probability that <math>B-J >2</math> is <math>1-\frac{1}{2}-\frac{1}{20} = \frac{9}{20} \implies \boxed{029}</math> | ||
− | |||
==Solution== | ==Solution== | ||
+ | By symmetry, the desired probability is equal to the probability that <math>J - B</math> is at most <math>-2</math>, which is <math>\frac{1-P}{2}</math> where <math>P</math> is the probability that <math>B</math> and <math>J</math> differ by <math>1</math> (no zero, because the two numbers are distinct). There are <math>20 * 19 = 380</math> total possible combinations of <math>B</math> and <math>J</math>, and <math>1 + 18 * 2 + 1 = 38</math> ones that form <math>P</math>, so <math>P = \frac{38}{380} = \frac{1}{10}</math>. Therefore the answer is <math>\frac{9}{20} \rightarrow \boxed{029}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | This problem is essentially asking how many ways there are to choose <math>2</math> distinct elements from a <math>20</math> element set such that no <math>2</math> elements are adjacent. Using the well-known formula <math>\dbinom{n-k+1}{k}</math>, there are <math>\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171</math> ways. Dividing <math>171</math> by <math>380</math>, our desired probability is <math>\frac{171}{380} = \frac{9}{20}</math>. Thus, our answer is <math>9+20=\boxed{029}</math>. | ||
+ | -Fidgetboss_4000 | ||
+ | |||
+ | ==Solution 3== | ||
+ | Create a grid using graph paper, with <math>20</math> columns for the values of <math>J</math> from <math>1</math> to <math>20</math> and <math>20</math> rows for the values of <math>B</math> from <math>1</math> to <math>20</math>. Since <math>B</math> cannot equal <math>J</math>, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since <math>B - J</math> must be at least <math>2</math>, we can mark the line where <math>B - J = 2</math>. Now we sum the number of squares that are on this line and below it. We get <math>171</math>. Then we find the number of total squares, which is <math>400 - 20 = 380</math>. Finally, we take the ratio <math>\frac{171}{380}</math>, which simplifies to <math>\frac{9}{20}</math>. Our answer is <math>9+20=\boxed{029}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | We can see that if <math>B</math> chooses <math>20</math>, <math>J</math> can choose from <math>1</math> through <math>18</math> such that <math>B-J\geq 2</math>. If <math>B</math> chooses <math>19</math>, <math>J</math> has choices <math>1</math>~<math>17</math>. By continuing this pattern, <math>B</math> will choose <math>3</math> and <math>J</math> will have <math>1</math> option. Summing up the total, we get <math>18+17+\cdots+1</math> as the total number of solutions. The total amount of choices is <math>20\times19</math> (B and J must choose different numbers), so the probability is <math>\frac{18*19/2}{20*19}=\frac{9}{20}</math>. Therefore, the answer is <math>9+20=\boxed{029}</math> | ||
+ | |||
+ | -eric2020 | ||
+ | |||
+ | ==Solution 5== | ||
+ | Similar to solution 4, we can go through the possible values of <math>J</math> to find all the values of <math>B</math> that makes <math>B-J\geq 2</math>. If <math>J</math> chooses <math>1</math>, then <math>B</math> can choose anything from <math>3</math> to <math>20</math>. If <math>J</math> chooses <math>2</math>, then <math>B</math> can choose anything from <math>4</math> to <math>20</math>. By continuing this pattern, we can see that there is <math>18+17+\cdots+1</math> possible solutions. The amount of solutions is, therefore, <math>\frac{18*19}{2}=171</math>. Now, because <math>B</math> and <math>J</math> must be different, we have <math>20\times19=380</math> possible choices, so the probability is <math>\frac{171}{380}=\frac{9}{20}</math>. Therefore, the final answer is <math>9+20=\boxed{029}</math> | ||
+ | |||
+ | -josephwidjaja | ||
+ | |||
+ | ==Video Solution == | ||
+ | https://www.youtube.com/watch?v=lh570eu8E0E | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/TSKcjht8Rfk?t=488 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | |||
+ | https://youtu.be/9X18wCiYw9M | ||
+ | |||
+ | ~Shreyas S | ||
+ | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=1|num-a=3}} | {{AIME box|year=2019|n=I|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:59, 6 October 2020
Contents
Problem 2
Jenn randomly chooses a number from . Bela then randomly chooses a number from distinct from . The value of is at least with a probability that can be expressed in the form where and are relatively prime positive integers. Find .
Solution
The probability that is by symmetry. The probability that is because there are 19 pairs: .
The probability that is
Solution
By symmetry, the desired probability is equal to the probability that is at most , which is where is the probability that and differ by (no zero, because the two numbers are distinct). There are total possible combinations of and , and ones that form , so . Therefore the answer is .
Solution 2
This problem is essentially asking how many ways there are to choose distinct elements from a element set such that no elements are adjacent. Using the well-known formula , there are ways. Dividing by , our desired probability is . Thus, our answer is . -Fidgetboss_4000
Solution 3
Create a grid using graph paper, with columns for the values of from to and rows for the values of from to . Since cannot equal , we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since must be at least , we can mark the line where . Now we sum the number of squares that are on this line and below it. We get . Then we find the number of total squares, which is . Finally, we take the ratio , which simplifies to . Our answer is .
Solution 4
We can see that if chooses , can choose from through such that . If chooses , has choices ~. By continuing this pattern, will choose and will have option. Summing up the total, we get as the total number of solutions. The total amount of choices is (B and J must choose different numbers), so the probability is . Therefore, the answer is
-eric2020
Solution 5
Similar to solution 4, we can go through the possible values of to find all the values of that makes . If chooses , then can choose anything from to . If chooses , then can choose anything from to . By continuing this pattern, we can see that there is possible solutions. The amount of solutions is, therefore, . Now, because and must be different, we have possible choices, so the probability is . Therefore, the final answer is
-josephwidjaja
Video Solution
https://www.youtube.com/watch?v=lh570eu8E0E
Video Solution 2
https://youtu.be/TSKcjht8Rfk?t=488
~IceMatrix
Video Solution 3
~Shreyas S
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.